Uniform convergence as $epsilonto 0^+$












1












$begingroup$


Reading some lectures on Hamilton-Jacobi PDE theory I found some terminology that I really don't understand.



Let $Omega$ be an open subset of $mathbb{R}^n$. Suppose that $u_epsilon:Omegato mathbb{R}$ is a function for $epsilon>0$. Consider a sequence of function $(u_epsilon)$ and let $u:Omegatomathbb{R}$.



What does $u_epsilon$ converges uniformly to $u$ as $epsilonto 0^+$ means?



I mean, I know what uniformly convergence means for a sequence of functions $(f_n)$ as $nto +infty$, but I really don't understand this terminology.



Thanks a lot in advance.






real-analysis uniform-convergence sequence-of-function







share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    Reading some lectures on Hamilton-Jacobi PDE theory I found some terminology that I really don't understand.



    Let $Omega$ be an open subset of $mathbb{R}^n$. Suppose that $u_epsilon:Omegato mathbb{R}$ is a function for $epsilon>0$. Consider a sequence of function $(u_epsilon)$ and let $u:Omegatomathbb{R}$.



    What does $u_epsilon$ converges uniformly to $u$ as $epsilonto 0^+$ means?



    I mean, I know what uniformly convergence means for a sequence of functions $(f_n)$ as $nto +infty$, but I really don't understand this terminology.



    Thanks a lot in advance.






    real-analysis uniform-convergence sequence-of-function







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      Reading some lectures on Hamilton-Jacobi PDE theory I found some terminology that I really don't understand.



      Let $Omega$ be an open subset of $mathbb{R}^n$. Suppose that $u_epsilon:Omegato mathbb{R}$ is a function for $epsilon>0$. Consider a sequence of function $(u_epsilon)$ and let $u:Omegatomathbb{R}$.



      What does $u_epsilon$ converges uniformly to $u$ as $epsilonto 0^+$ means?



      I mean, I know what uniformly convergence means for a sequence of functions $(f_n)$ as $nto +infty$, but I really don't understand this terminology.



      Thanks a lot in advance.






      real-analysis uniform-convergence sequence-of-function







      share|cite|improve this question









      $endgroup$




      Reading some lectures on Hamilton-Jacobi PDE theory I found some terminology that I really don't understand.



      Let $Omega$ be an open subset of $mathbb{R}^n$. Suppose that $u_epsilon:Omegato mathbb{R}$ is a function for $epsilon>0$. Consider a sequence of function $(u_epsilon)$ and let $u:Omegatomathbb{R}$.



      What does $u_epsilon$ converges uniformly to $u$ as $epsilonto 0^+$ means?



      I mean, I know what uniformly convergence means for a sequence of functions $(f_n)$ as $nto +infty$, but I really don't understand this terminology.



      Thanks a lot in advance.






      real-analysis uniform-convergence sequence-of-function




      real-analysis uniform-convergence sequence-of-function






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 12 at 16:46









      eleguitareleguitar

      124114




      124114






















          1 Answer
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          $begingroup$

          Let ${u_h}_h$ be a family of functions indexed by $hin Ssubset (0,infty)$. We say that $u_hto u$ uniformly as $hto 0^+$ if for a given $varepsilon>0$, we can find $delta>0$ such that
          $$
          sup_{xinOmega}|u_h(x)-u(x)| < varepsilon
          $$
          for all $hin Scap(0,delta)$.




          Remark: Using this definition, we can see that $u_nto u$ uniformly in the usual sense iff $hat u_hto u$ uniformly as $hto 0^+$, provided that we take our index set to be $S={1,frac 12,frac 13,dots}$ and
          $$
          hat u_{1/n} := u_n.
          $$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            That's completely clear! Thank you so much!
            $endgroup$
            – eleguitar
            Jan 12 at 18:45











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          1 Answer
          1






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          active

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          2












          $begingroup$

          Let ${u_h}_h$ be a family of functions indexed by $hin Ssubset (0,infty)$. We say that $u_hto u$ uniformly as $hto 0^+$ if for a given $varepsilon>0$, we can find $delta>0$ such that
          $$
          sup_{xinOmega}|u_h(x)-u(x)| < varepsilon
          $$
          for all $hin Scap(0,delta)$.




          Remark: Using this definition, we can see that $u_nto u$ uniformly in the usual sense iff $hat u_hto u$ uniformly as $hto 0^+$, provided that we take our index set to be $S={1,frac 12,frac 13,dots}$ and
          $$
          hat u_{1/n} := u_n.
          $$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            That's completely clear! Thank you so much!
            $endgroup$
            – eleguitar
            Jan 12 at 18:45
















          2












          $begingroup$

          Let ${u_h}_h$ be a family of functions indexed by $hin Ssubset (0,infty)$. We say that $u_hto u$ uniformly as $hto 0^+$ if for a given $varepsilon>0$, we can find $delta>0$ such that
          $$
          sup_{xinOmega}|u_h(x)-u(x)| < varepsilon
          $$
          for all $hin Scap(0,delta)$.




          Remark: Using this definition, we can see that $u_nto u$ uniformly in the usual sense iff $hat u_hto u$ uniformly as $hto 0^+$, provided that we take our index set to be $S={1,frac 12,frac 13,dots}$ and
          $$
          hat u_{1/n} := u_n.
          $$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            That's completely clear! Thank you so much!
            $endgroup$
            – eleguitar
            Jan 12 at 18:45














          2












          2








          2





          $begingroup$

          Let ${u_h}_h$ be a family of functions indexed by $hin Ssubset (0,infty)$. We say that $u_hto u$ uniformly as $hto 0^+$ if for a given $varepsilon>0$, we can find $delta>0$ such that
          $$
          sup_{xinOmega}|u_h(x)-u(x)| < varepsilon
          $$
          for all $hin Scap(0,delta)$.




          Remark: Using this definition, we can see that $u_nto u$ uniformly in the usual sense iff $hat u_hto u$ uniformly as $hto 0^+$, provided that we take our index set to be $S={1,frac 12,frac 13,dots}$ and
          $$
          hat u_{1/n} := u_n.
          $$






          share|cite|improve this answer











          $endgroup$



          Let ${u_h}_h$ be a family of functions indexed by $hin Ssubset (0,infty)$. We say that $u_hto u$ uniformly as $hto 0^+$ if for a given $varepsilon>0$, we can find $delta>0$ such that
          $$
          sup_{xinOmega}|u_h(x)-u(x)| < varepsilon
          $$
          for all $hin Scap(0,delta)$.




          Remark: Using this definition, we can see that $u_nto u$ uniformly in the usual sense iff $hat u_hto u$ uniformly as $hto 0^+$, provided that we take our index set to be $S={1,frac 12,frac 13,dots}$ and
          $$
          hat u_{1/n} := u_n.
          $$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 12 at 17:06

























          answered Jan 12 at 17:00









          BigbearZzzBigbearZzz

          8,69121652




          8,69121652












          • $begingroup$
            That's completely clear! Thank you so much!
            $endgroup$
            – eleguitar
            Jan 12 at 18:45


















          • $begingroup$
            That's completely clear! Thank you so much!
            $endgroup$
            – eleguitar
            Jan 12 at 18:45
















          $begingroup$
          That's completely clear! Thank you so much!
          $endgroup$
          – eleguitar
          Jan 12 at 18:45




          $begingroup$
          That's completely clear! Thank you so much!
          $endgroup$
          – eleguitar
          Jan 12 at 18:45


















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